AMPLIFIER Circuits [Transistor Circuits (1964)]

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An amplifier receives a signal and "boosts" it to a higher level for application to another amplifier or to an output device.

The operation of four types of amplifier circuits is discussed in this Section. All are audio amplifiers; however the same basic principles apply to RF amplifiers except, of course, the component values are different and transformer coupling is usually employed.

CLASS-A AUDIO AMPLIFIER

Figs. 1 and 2 show two successive half-cycles in the operation of a typical audio-amplifier circuit which utilizes trans former coupling to the next stage. The circuit is classified as a Class-A amplifier because the collector-emitter current flows continuously during the entire cycle.

Identification of Components

The following components perform necessary functions in this amplifier circuit:

R1-Voltage divider and biasing resistor.

R2-Voltage divider and biasing resistor.

R3-Emitter stabilizing resistor.

C1-Input coupling and blocking capacitor.

C2-Emitter bypass or filter capacitor.

T1-Audio-frequency output transformer.

X1-NPN transistor.

M1-Battery or other DC source.


Fig. 1. Operation of the Class-A amplifier-negative half-cycle.


Fig. 2. Operation of the Class-A amplifier-positive half-cycle.

Identification of Currents

A total of six different electron currents perform the various essential functions in this circuit. The currents are:

1. Voltage-divider current ( dotted green).

2. Base-emitter current (solid green).

3. Collector-emitter current (frequently referred to as the collector current; solid red).

4. Input signal current (solid blue).

5. Output signal current (dotted blue).

6. Emitter filter current (dotted red).

The first three are static currents, which begin to flow as soon as power is applied to the circuit. In the absence of an applied signal voltage or current, these three static currents will be essentially pure DC. When a signal voltage is applied to the circuit to be amplified, the base-emitter and collector emitter currents will become pulsating DC. As soon as a signal voltage is applied, the circuit is then operating under dynamic conditions, and the latter three currents will come into existence.

Details of Operation

The current through resistors R1 and R2 (shown in dotted green) flows clockwise into the positive terminal of battery M1 and out the negative terminal, returning to the lower terminal of R1. The use of two or more resistors across a fixed voltage source is a very common means of obtaining some fraction of the total voltage available. It is a particularly valuable technique in transistor circuits, which require only 1 or 2 volts for biasing the base with respect to the emitter, and another volt age perhaps three to five times as large-but always of the same polarity--for biasing the collector with respect to the emitter.

These requirements can of course be met by two separate voltage sources or batteries, but it is usually desirable to have as few batteries as possible in any circuit. Since the current drain on batteries in transistor circuits is normally very small, the power losses incurred by a voltage-divider current such as this one are well compensated for by the elimination of a separate battery for providing low-voltage bias between base and emitter.

The base-emitter current flowing through a transistor connected like this one will normally be only 10 or 20 microamperes.

This current (in solid green) flows upward from ground through resistor R3 and through the NPN transistor, from emitter to base. It then goes to the right, down through resistor R2, and enters the positive terminal of M1. The current then flows through the battery and out its negative terminal, to common ground.

The collector-emitter current flows in a clockwise direction through the closed loop which starts at ground, then upward through resistor R3, through the transistor from emitter to collector, downward through the primary winding of transformer T1, and into the positive terminal of battery M1. Like the two other static currents, this one must also flow through the battery, from the positive to the negative terminal (ground). This circuit does not function as an amplifier until a signal voltage is applied to its input circuit. Its purpose is to amplify, or increase the strength of, this signal voltage. Let us consider now how this important function is accomplished. An alternating signal voltage at an audio frequency will be applied to input capacitor C1. The result is that a signal current (shown in solid blue) will be made to flow, at the same frequency, up and down through input driving resistor R1. Fig. 1 has been labeled the negative half-cycle, because this input signal current flows downward through R1 and develops a small component of negative voltage across this resistor. This negative component must be subtracted from the positive voltage developed across this same resistor by the voltage-divider current. The net result, during the negative half-cycle shown in Fig. 1 is a reduction in the positive voltage applied to the base of the transistor. Since this is an NPN, less electron current flows from emitter to base.

Probably the most important single truth in the internal physics of the transistor action is that a slight increase ( or decrease) in the emitter-to-base (or base-to-emitter) current will produce a much larger increase ( or decrease) in the emitter to-collector ( or collector-to-emitter) current. This is the central fact that enables the transistor to be used as an amplifier.

In Fig. 1, the voltage to be amplified is negative and there fore drives a signal current (solid blue) downward through R1, making the top of this resistor negative. As a result of this negative biasing action at the base, the main electron stream through the transistor (the collector-emitter current) is reduced to its minimum value. This decrease in current flowing downward through the primary of output transformer T1 causes the current to flow in the same direction in the secondary winding. This is normal transformer action at work, and it accounts for the out put current (in dotted blue) flowing downward in the secondary in Fig. 1.

In Fig. 2, the phase of the input signal voltage is reversed.

Being positive, it now draws the signal-current electrons upward through resistor R1, adding to the positive voltage already created there by the upward flow of voltage-divider current (in dotted green). The higher positive voltage increases the flow of both transistor currents; and as more collector current flows down ward through the primary, more output current flows upward in the secondary.

The transformer action is probably the most difficult of the five basic electronic actions to visualize. When an alternating current is driven through either winding, it becomes the primary winding, and the current through it can be labeled the primary current. The primary current will cause a secondary current to flow in the other winding. The phase relationship between the two currents will always be governed by the following considerations:

1. When the primary current is increasing in the downward direction, the secondary current must be decreasing in the downward direction, or increasing in the upward direction.

2. When the primary current is decreasing in the downward direction, the secondary current must be increasing in the downward direction, or decreasing in the upward direction.

Obviously, two similar rules can be written to account for increases or decreases in the primary current when flowing upward.

The quantity relationship between primary and secondary currents is regulated by the turns ratio of the transformer.

Expressed as a formula:

where,

N1 is the number of turns of wire in the primary, N2 is the number of turns of wire in the secondary, I1 is the quantity of primary current, I2 is the quantity of secondary current,

E1 is the voltage impressed across the primary winding,

E2 is the voltage induced across the secondary winding.

In order for the circuit as a whole to function as a voltage amplifier, the alternating instantaneous voltage E1 across the primary winding must be greater than the instantaneous voltage developed across R1 by the input-signal current.

The classification of this circuit as a Class-A amplifier tells us that the collector-emitter current is never completely cut off during the entire cycle; some current flows continuously through the primary of T1.

Degeneration, or loss of signal strength, is prevented in this circuit by the presence of filter capacitor C2 across emitter resistor R3. C2 keeps the pulsations in collector-emitter cur rent from developing voltage pulsations across R3. During the positive half-cycle of Fig. 2, the biasing conditions are such that they encourage or demand more collector-emitter current through the transistor. These extra electrons are drawn from the upper plate of capacitor C2, rather than through resistor R3.

An equal number of electrons (the filter current, in dotted red) are drawn onto the lower plate of C2 from ground.

The opposite action occurs during the negative half-cycle of Fig. 1. Now biasing conditions are such that they discourage or restrict the flow of electrons which make up the collector emitter current. The electrons flowing upward through R3 are momentarily shunted aside and flow onto the upper plate of capacitor C2, driving an equal number of filter-current electrons downward from the lower plate of C2 to ground.

Thus, C2 is carrying out its normal capacitor action of "passing" an alternating current from one plate to another without permitting any continuous flow in a single direction (DC). The values of R3 and C2 must be so chosen that the resistance of R3 is much greater than the capacitive reactance (opposition to alternating current flow) of C2. The reactance of any capacitor varies inversely with the frequency of the current being passed, in accordance with the standard formula:

1 X('=- 21rfC

.... where, Xe is the reactance in ohms, f is the frequency in cycles per second, C is the capacitance in farads.

When this gross inequality between resistance ( of R3 ) and reactance (of C2 ) has been met, then the combination of the two components automatically forms a "long time-constant" circuit.

The product of the resistance in ohms and the capacitance in farads equals the time constant of the circuit, in accordance with the formula: T=RxC where, T is the time constant in seconds, R is the resistance in ohms, C is the capacitance in farads.

Any combination where the time constant, T, is at least five times longer than the time required for a single cycle of the current being passed to complete itself is defined as a long time-constant circuit. The time for a single cycle of any current is related to the frequency of that current by the simple relationship: where, T is the time duration of a single cycle, f is the frequency in cycles per second.

The voltage which accumulates on the upper plate of capacitor C2 may be likened to a deep pool of positive ions, out of which electrons are drawn during the positive half-cycle of Fig. 2, and to which electrons are added during the negative half-cycle of Fig. 1. The true significance of a "long time-constant" RC circuit is that not enough electrons are withdrawn or added to appreciably change the voltage represented by this ion pool.

Another way of saying this is that so few electrons are with drawn during a positive half-cycle, in comparison with the number of positive ions already stored there, that no measurable increase in positive voltage occurs. Likewise, the number of negative electrons added to this ion pool during a negative half-cycle is so small, in comparison with the number of positive ions already stored there, that no measurable decrease in positive voltage occurs.

There is a simple arithmetical relationship between the quantity of electrons (or ions) stored on a capacitor plate, the size of the capacitor, and the resulting voltage across the capacitor. Known as Coulomb's law, it is written as follows:

Q=CxE

…where, Q is the quantity of charge in Coulombs ( one coulomb is equal to 6.25 x 1018 negative electrons or positive ions), C is the size of the capacitor in farads, E is the resulting voltage in volts.

DIRECT-COUPLED AMPLIFIER

Figs. 3 and 4 show two successive half-cycles in the operation of a direct-coupled (DC) amplifier using transistors.

Direct coupling is advantageous in that the inherent losses in capacitive or transformer coupling, at low and high frequencies, respectively, are avoided.

The principle known as complementary symmetry is used in this circuit. Here the collector of an NPN transistor, X1, has been coupled directly to the base of a PNP transistor, X1. The significance of the biasing-voltage polarities and current-flow directions in the two types of transistors will be discussed later in the Section.

Identification of Components

This circuit is composed of the following components:

R1-Input driving and voltage-dividing resistor.

R2-Voltage-dividing resistor.

R3-Emitter stabilizing resistor.

R4-Collector load resistor.

R5--Emitter stabilizing resistor.

RS-Collector load resistor.

C1-Input capacitor for coupling signal from preceding stage.

C2-Emitter bypass capacitor.

C3-Emitter bypass capacitor.

X1-NPN transistor.

X2-PNP transistor.

M1-Battery or other DC power supply.

Identification of Currents

There are at least eight significant electron currents at work in this circuit. Their movements and interrelationships must be understood by anyone aspiring to know how this circuit operates. These currents are:

1. Input dividing current ( dotted blue) .

2. Voltage-divider current (dotted green).

3. Base-emitter current for transistor X1 (solid green).

4. Collector-emitter current for transistor X1 (solid red).

5. Base-emitter current for transistor X2 (solid blue).

6. Collector-emitter current for transistor X2 ( dotted red).

7. Emitter filter current for transistor X1 (also in solid red).

8. Emitter filter current for transistor X2 (also in dotted red).

Details of Operation Fig. 3 has been labeled a negative half-cycle of operation, because the input driving signal flows downward through resistor R1, making the voltage at the base of transistor X1 more negative during this half-cycle. Fig. 4 has been labeled a positive half cycle because now the input driving signal flows upward through R1, making the base of X1 more positive. The instantaneous volt ages which this input-signal current develops at the top of R1 must be added to or subtracted from the more permanent positive voltage developed there by the voltage-divider current (in dotted green).

This voltage-divider current flows through the closed path from ground, upward through R1, across and down through R2, and then into the positive terminal of power supply M1. From here, it flows through the battery and out the negative terminal to the common-ground connection. The purpose of this current (and consequently, of the voltage divider itself) is to develop a particular value of positive voltage at the junction of R1 and R2. This becomes the biasing voltage for the base of transistor X1. The term "biasing voltage" requires some explanation.

Transistors are normally considered "current-controlled" devices, and the base-emitter current is usually referred to as the "biasing current." However, no biasing current will flow unless certain values of voltage are applied to the base and emitter. It is quite proper to refer to these values as "biasing voltages," because they determine the amount of biasing current which will flow between base and emitter.

The actual biasing current flowing from emitter to base has been shown in solid green. Its electrons flow upward from ground, through emitter stabilizing resistor R3, into the transistor (flowing against the direction of the emitter arrow, of course), then out of the base and downward through resistor R2 to the positive terminal of battery M1. When no signal is applied to input capacitor C1 to be amplified, this biasing current is a pure DC. In the negative half-cycle of Fig. 3, the negative voltage which the downward-flowing signal current develops at the top of R1 will reduce the biasing current through transistor X1.

The lower biasing current will in turn reduce the collector cur rent (in solid red). The complete path of this emitter-collector current begins at ground, below resistor R3. The current flows upward, through R3, into the emitter and out the collector of X1, and down through load resistor R4 to the positive terminal of the battery. This electron current then returns to common ground by flowing through the battery and out its negative terminal.

At that moment when the signal current in the negative half cycle of Fig. 3 is flowing downward through R1, at its maximum rate, the biasing and collector currents will have their minimum values. The reduction in collector current causes the voltage at the collector to become more positive. The reason is this voltage will always be equal to the power-supply voltage (positive in this case) minus the voltage drop occasioned across R4 by the flow of collector-emitter current through it. As this current decreases, so does the resulting voltage drop across R4, and the voltage at the collector will rise toward the full value of power~ supply voltage.


Fig. 3. Operation of the direct-coupled amplifier-negative half-cycle.


Fig. 4. Operation of the direct-coupled amplifier-positive half-cycle.

The exception to the foregoing would be where the collector emitter current is cut off entirely. Then the voltage difference between the two terminals of R4 ( the voltage drop across R4) would be zero, and the collector voltage would necessarily be the same as the power-supply voltage.

Since the collector of X1 is coupled or connected directly to the base of X2, the voltages at these two elements must always be identical. Let us now consider the biasing conditions at the elements of X2, so that we may then predict what will happen to the currents through X2 during this negative half-cycle.

Since X2, is a PNP transistor, electron currents will flow into its collector and out its emitter-instead of from emitter to collector as in NPN transistor X1.

The important biasing voltages for X2 (or any other transistor) are the instantaneous voltages at the base and emitter. You have already seen that the voltage at the base of X2 is determined at all times by the voltage at the collector of X1. The voltage at the emitter of X2 is jointly determined by the power-supply volt age of M1 and by the voltage drop across emitter resistor R5 as the collector-emitter current flows through R5. The complete path of this collector current (in dotted red) begins at ground, below resistor R6. It flows upward through R6 into the collector and out the emitter of the transistor, then down through R5 to the positive terminal of power supply M1. Its journey is completed to common ground by flowing through the battery and out its negative terminal.

The voltage which this current flow develops across emitter resistor R5 must be subtracted from the battery voltage to determine the exact voltage at the emitter (the emitter biasing voltage). It and the voltage at the base (the base biasing voltage previously discussed) regulate the amount of biasing current flowing from base to emitter. As with all transistors, the amount of biasing current through the transistor exercises direct control over the amount of collector-emitter current. This brings us back to a consideration of how much voltage this collector current will develop across R5 while flowing through it; this voltage affects the emitter biasing voltage, etc.

The complete path of the emitter-base current (in solid blue) truly begins at ground, below transistor X1. It flows through resistor R3 and transistor X1 as part of the collector current (in solid red). At the collector of X1 the emitter-base current separates from the main collector current and assumes its individual identity, flowing directly into the base of transistor X1 and out its emitter, then downward through resistor R5 to the positive terminal of battery M1. From here it can return to ground by flowing through the battery.

During the negative hall-cycle shown in Fig 3 (collector current through X1 reduced by the biasing conditions at X1), the positive collector voltage is increased and so is the positive voltage at the base of X2. The latter action reduces the base emitter current (biasing current) through X2. It may be difficult to visualize why a more positive base voltage for X2 reduces its base-emitter current. If so, consider again the exception for transistor X1, when we assumed that no collector current at all was flowing through X1. This no-current condition raised the voltage at the collector of X1 until it equaled the power-supply voltage. The base of X2 would necessarily assume the same voltage value. Obviously, no current would then flow between the base and the positive terminal of the battery, since the two points would be at the same voltage. (No electron current can flow between two points unless a difference in voltage exists between them. Electrons will always flow from the more negative point (point of greater electrons) to the more positive point (point of fewer electrons) to make up for the deficiency there.) During the positive half-cycle depicted in Fig. 4, the following changes in voltage polarities and current quantities occur:

1. The signal current ( dotted blue) flows upward through R1, making the voltage at the base of X1 more positive.

2. The base-emitter biasing current (solid green) through X1 increases.

3. The emitter-collector current (solid red) through X1 also increases.

4. The positive voltage at the collector of X1 decreases and lowers the positive voltage at the base of X2.

5. The base-emitter biasing current (solid blue) through X2 increases.

6. The collector-emitter current (solid blue) through X2 also increases.

Comparison of Output and Input Voltages

The output voltage for this amplifier circuit is taken from the collector of X2. It will vary from a low to a high positive value, depending on the amount of collector current. During the negative half-cycle shown in Fig. 3, the collector current is reduced to its minimum value; therefore, the voltage it develops at the top of load resistor R6 will have its lowest positive value, too. This collector voltage will be positive, because the collector current always flows upward through R6, and electron current always flows from a less positive to a more positive point.

During the positive half-cycle of Fig. 4, the collector cur rent reaches its maximum value; consequently, the instantaneous output voltage at the top of R6 will have its maximum positive value. The magnitude of the difference in the two collector volt ages is the peak-to-peak value of the amplified voltage. With the circuit shown here, this peak-to-peak output might easily reach 12 or 15 volts, even though the input signal voltage developed across resistor R1 by the signal current will normally have a peak to peak value of only a small fraction of a volt.

Therefore, we can say that the input-signal voltage has been substantially "amplified." The output voltage will be "in phase" with the input-signal voltage. This means that in the negative half-cycle of Fig. 3 (input voltage negative at the top of resistor R1), the output voltage at the top of resistor R6 will have its least positive value. Conversely, when the input voltage is positive (as in Fig. 4), the output voltage will have its highest positive value.

Filter Currents

The emitter resistors, R3, and R5, are bypassed to ground by C2 and C3. These two filter capacitors prevent loss of signal voltage due to degeneration. (Degeneration was described at some length in the previous amplifier discussion.) Without going into extensive detail about the various causative factors, the significant and observable effects that occur in capacitor C2 and resistor R3 can be tabulated:

1. The currents through transistor X1 flow upward through R3 therefore the voltage at the top of R3 has to be positive. The upper plate of capacitor C2 will always assume the same voltage, which can be likened to a pool of positive ions.

2. The combination of R3 and C2 forms a "long time-constant" circuit. Consequently, the pool of positive ions on the upper plate is so overwhelmingly large, in comparison with the number of electrons drawn out of it and into the transistor during the positive half-cycles, that the positive voltage on the plate does not increase during these half-cycles.

3. This pool of positive ions is also overwhelmingly large, in comparison with the number of electrons which flow onto the upper plate of C2 from the top of resistor R3 during the negative half-cycles. Therefore, the positive voltage on this plate does not decrease during these negative half-cycles.

4. When the collector current decreases during the negative half-cycles (Fig. 3) electrons flow onto the upper plate of C2, driving the filter current downward from the lower plate of C2 to ground.

5. When the collector current increases (Fig. 4), electrons are drawn from the upper plate of C2. A like number of electrons are drawn upward from ground onto the lower plate of C2.

When all of these actions are permitted to occur, the emitter resistor is said to be bypassed or filtered, and degeneration has been avoided. Otherwise, the following would occur:

1. On negative half-cycles, the positive voltage at the emitter would fall as the collector current through R3 is reduced.

The base-emitter and collector-emitter currents would tend to increase and thus nullify part of the original decrease in collector current.

2. Degeneration during positive half-cycles would be characterized by a rise in the positive voltage at the emitter as the collector current increases. Now the base-emitter and collector-emitter currents would tend to decrease and thus nullify part of the original increase in collector current.

The filtering action occurring across capacitor C3 is identical in nature but opposite in phase to that just described for C2.

A decrease in collector current through X2 during negative half cycles causes filter current to flow out of ground and onto the bottom plate of C3. An increase in collector current through X2 during positive half-cycles would drive this filter current back into ground.

Complementary Symmetry

The term "complementary symmetry" refers to the fact that currents flow in opposite directions in an NPN and a PNP transistor. Also, it refers to the fact that to increase conduction in an NPN transistor, the signal applied to its base must be positive; whereas in the PNP transistor, the signal applied to its base must be negative.

The principle of complementary symmetry will be utilized later in this Section in a push-pull circuit.

RC-COUPLED AMPLIFIER WITH NEGATIVE FEEDBACK

Figs. 5 and 6 show two successive half-cycles in the operation of a two-transistor audio amplifier which employs RC coupling and a negative-feedback network. The important ad vantage of negative feedback is a reduction in distortion. Other advantages are reduction in the variation in gain provided by different transistors, and an apparent increase in input impedance of the circuit using the feedback. The inevitable dis-advantage of negative feedback is some loss in gain-however, this is a small price to pay for the many advantages obtained through its use.

Identification of Components

The following circuit components perform the functions indicated. The manner in which these functions are accomplished will be elaborated on when the individual electron currents are discussed.

R1-Input driving resistor.

R2-Emitter stabilizing resistor for X1.

R3-Collector load resistor for X1.

R4-Voltage-dividing and biasing resistor.

R5-Voltage-dividing and biasing resistor.

R6-lnput driving resistor for X2.

R7-Emitter stabilizing resistor for X2.

R8-Collector load resistor for X2.

R9-Feedback resistor.

C1-Input coupling capacitor.

C2-lnterstage coupling capacitor.

C3-Output coupling capacitor.

C4-Feedback capacitor.

C5-Emitter filter capacitor.

C6-Voltage-divider filter capacitor.

X1 and X2-PNP transistors.

M1-Battery power supply.

Identification of Currents

This circuit has at least ten separate electron currents at work during normal operation. To understand how such a circuit works, you must be able to visualize each current-what makes it flow and what this flow in turn accomplishes, what its complete flow path is, etc. Once the currents are understood and visualized, the significance of their various functions will be perfectly clear.

1. Input driving current (solid blue).

2. Voltage-divider current (dotted green).

3. Base-emitter current through each transistor (solid green).

4. Collector-emitter current through X1 (solid red).

5. Driving current for transistor X2 ( also in solid blue) .

6. Collector-emitter current for X2 ( dotted red).

7. Feedback current (dotted blue).

8. Emitter filter current for X2 (also in dotted red).

9. Output current ( also in dotted blue) .

10. Filter currents across part of voltage divider (also m dotted blue).

Details of Operation

As long as no signal is applied to the input capacitor, the driving current shown in solid blue will not exist, and the rest of the circuit will be operating under a static condition. Let us consider first the five currents which flow during such a static condition-namely, the voltage-divider current, and the two currents through each transistor.

The voltage-divider current (in dotted green) flows continuously in a clockwise path, upward through resistors R5 and R4 and downward through battery M1 from its positive to its negative terminal. The flow of this current through R5 develops a positive voltage at the top of R5, and this voltage is applied directly to the base of each transistor through another set of resistors. Consequently, the positive voltage at the top of R5 can be labeled the "base biasing voltage." Since the full battery voltage (22.5 volts) exists across both of the resistors in series, the amount across R5 can be determined by a simple proportional relationship involving the sizes of R5 and R4 and the battery voltage. The formula is:

E R5 E u5=R.l+R5 X :111

…where, Eur, is the voltage developed across R5 by the voltage-divider currents, in volts, R5 is the resistance of R5 in ohms, R4 is the resistance of R4 in ohms, E111 is the voltage of the battery in volts.

The base-emitter current through X1 (solid red) will flow initially in an amount determined by the voltage at the base and emitter. This current flow through emitter resistor R2 develops a small positive voltage at the top of R2, and it may be labeled the "emitter biasing voltage." Flowing downward through resistor R1, this current develops a small component of negative voltage at the top of R1, and this component reduces somewhat the positive voltage applied to the base from the voltage divider. The complete path of this electron current is upward from ground through resistor R2, through the transistor from emitter to base, and downward through R1 to the junction of R4 and R5. Then it goes upward, through R4, to the positive terminal of battery M1, where the chemical action of the battery makes it flow out the negative terminal and back to ground. The reason this electron current flows upward rather than downward at the junction of R4 and R5 is the attraction of the positive terminal of the battery. This is the highest positive voltage in the loop made by the current, and consequently the point toward which all electron currents will be drawn.

Since transistors are "current-operated" devices, the flow of an electron current from emitter to base will change the cur rent flow from emitter to collector. (The emitter-to-base current is called the biasing current.) The upward flow of collector cur rent through resistor R2 will develop an additional component of positive voltage at the top of R2. This voltage further modifies the emitter biasing voltage, and consequently affects the total amount of biasing current which will flow through the transistor from emitter to base.

The complete path of the collector current begins at ground, below R2, and flows upward through R2 into the emitter and out the collector. Then it goes downward through the collector load resistor R3, into the positive terminal of battery M1, and returns to ground from the negative terminal.

Within transistor X2, an initial flow of electron current from emitter to base is set in motion by application of the base biasing voltage obtained from the junction of R4 and R5 as a result of the voltage-divider current action previously discussed. Its complete path, shown in solid green is upward from ground through R7, into the emitter and out the base, downward through R6, then upward through R4 to the positive terminal of battery M1. As soon as this initial current begins to flow, it will alter both the emitter and the base voltage. The voltage at the top of R7 will become more positive, and the voltage at the top of R6 will become more negative. Both voltage changes are of such polarity that they restrict or oppose the flow of the base-emitter current.


Fig. 5. Operation of the resistance-capacitance coupled amplifier with negative feedback-negative half-cycle.

Transistors are very sensitive to temperature changes-a rise in temperature causes more current to flow between emitter and base. In turn, more emitter-collector current flows, the transistor becomes still hotter and the current further increases.

Such runaway condition will eventually destroy the transistor.

The presence of emitter stabilizing resistors such as R2 and R7 prevents this from happening. Since all the current for X2 must first flow upward through R7, these current increases will rapidly make the voltage at the top of R7 so positive that. it will tend to restrict or oppose the flow of these currents. (Remember that the voltage applied to an emitter is one of the two important biasing voltages of a transistor.) The emitter-collector current for X2 (in dotted red) flows upward from ground, through R7, into the emitter and out the collector. Then it heads downward through load resistor RS and into the positive terminal of the battery.

In summary, it can be verified that all four of the currents which flow through the two transistors are drawn from ground and through their respective paths by the positive voltage of the power supply or battery.


Fig. 6. Operation of the resistance-capacitance coupled amplifier with negative feedback-positive half-cycle.

Operation Under Dynamic Conditions

The application of a signal voltage to input capacitor C1 brings six more electron currents into existence, and all of them will be true alternating currents-meaning that they will periodically reverse their directions of flow along their respective paths in accordance with the frequency of the applied signal.

Additionally, the presence of an input signal will modify the four currents flowing through the two transistors, so that they will become pulsating rather than pure direct currents.

When an alternating voltage to be amplified is applied to input capacitor C1 the voltage begins to drive an electron cur rent up and down through resistors R1 and R5. Fig. 5 has been labeled the negative half-cycle because the input voltage is negative during this period. This negative voltage causes the input driving current (in solid blue) to flow downward through R1 and R5 to ground. The component of negative voltage now developed at the top of R1 must be subtracted from the positive voltage already applied there by the voltage-divider action at the junction of R4 and R5. Since the upper terminal of R1 is connected directly to the base of transistor X1, this over-all decrease in positive voltage at the base during the negative half-cycle will oppose and thereby restrict the flow of emitter-base cur rent ( in solid green). Any such reduction in the emitter-base biasing current will cause a much larger reduction in the emitter-collector current through the transistor. The latter may be looked upon as the main electron stream through the transistor. A reduction in this cur rent, as it flows through load resistor R3, causes the positive voltage at the top of R3 (and consequently at the collector) to rise. To satisfy yourself that a reduction in collector current causes a rise in collector voltage, suppose the collector current were cut off entirely. Under this condition, there could be no voltage drop whatsoever across R3, so both of its terminals would necessarily assume the full positive voltage of the battery or power supply.

When the collector voltage rises during the negative half cycles, electrons will be drawn toward this point from any external circuit connected to it. This accounts for the upward movement of the driving current (in solid blue) through biasing and driving resistor R6. The component of positive voltage now created at the top of R6 adds to the positive biasing voltage already existing at the base of X2 as a result of the voltage divider current through R5 and R4.

The resulting increase in positive voltage at the base of X2 acts to encourage, or increase, the flow of emitter-base biasing current through X2 (in solid green). In turn, the emitter collector current also increases and, in so doing, creates a larger voltage drop across load resistor RS. The amount of this voltage drop must of course be subtracted from the power-supply voltage to determine the instantaneous value of collector voltage.

Thus, you can see that as the collector current rises, the positive collector voltage drops in value.

This drop in collector voltage is simultaneously "passed" to the output circuit through capacitor C3 and, via the feedback network, back to the emitter of transistor X1. Physically this is accomplished by electrons being driven into each external circuit, as shown in Fig. 5. This occurs because excess electrons are momentarily pouring out of the collector but cannot immediately flow downward through load resistor RS; therefore they choose any alternate path available.

Consequently, during this negative half-cycle a feedback current (in dotted blue) flows to the left, through resistor R9, and into feedback capacitor C4. An equal number of electrons are driven away from the left plate of C4 and downward through emitter resistor R2. The downward movement through R2 places a small component of negative voltage at the emitter of X1, which must be subtracted from the positive voltage normally existing there as a result of the two transistor currents flowing upward through R2. Any reduction in the positive voltage at this point (previously identified as the "emitter biasing voltage") will increase the electron current flowing from emitter to base through the transistor (the "biasing" current of the transistor). Since the negative portion of the signal had originally reduced the biasing current through transistor X1, this feedback action must be classified as negative because it opposes the signal action and thereby lowers the total gain available from the transistor.

Recall earlier that when the negative portion of the signal was applied to the base, less collector current flowed. As a result, an entirely independent negative-feedback action occurs simultaneously across emitter resistor R2. The decrease in both the emitter and collector currents flowing upward through R2 lowers the voltage drop across R2 and also reduces the positive voltage at the emitter (the emitter biasing voltage). This change in the bias conditions of the transistor will tend to increase the biasing current flowing from emitter to base. Thus the biasing current is acting in opposition to the negative signal at the base because, during this same negative half-cycle, the signal is trying to reduce the flow of this current through the transistor.

The common name for this particular negative feedback is degeneration. One drawback is that it reduces the gain available from the transistor.

Actions During Positive Half-Cycle

During the positive half-cycle shown in Fig. 6, the following changes in current directions and voltage polarities occur:

1. The signal current (in solid blue) flows upward through R1, developing a component of positive voltage at the base.

As a result, more emitter-base biasing current (in solid green) flows through the transistor.

2. The emitter-collector current (in solid red) is correspondingly increased, causing a driving current (in solid blue) to flow downward through resistor R6. The latter develops a small component of negative voltage which alters the bias of transistor X2 in such a manner that less base-emitter biasing current now flows through the transistor.

3. In turn, the collector-emitter current through X2 also de creases, and as it falls, the positive voltage at the collector rises.

4. The higher positive voltage draws electrons toward the collector from any external circuit connected to it. This ac counts for the reversal in flow of the output and feedback currents.

5. Since feedback current is drawn toward the collector, it must flow upward through emitter resistor R2. As it does, it creates a small component of positive voltage at the top of R2. Now the emitter biasing conditions of X1 are altered in such a manner that less base-emitter biasing current flows through X1. Since the positive portion of the applied signal is simultaneously trying to increase this biasing current, the feedback can again be identified as negative.

6. Negative feedback known as degeneration also occurs across resistor R2 during this half-cycle. The signal increases the two currents through transistor X1, but in flowing up ward through R2, the now higher currents will increase the positive voltage at the emitter. In an NPN transistor, a more positive emitter will reduce the currents through the transistor.

Filter Current

Emitter resistor R7 for transistor X2 is bypassed with filter capacitor C5, so that degeneration does not occur across R7. The fluctuations from half-cycle to half-cycle in the current going into the emitter of X2 are in a sense "absorbed" by the filter capacitor. As a result, a constant current flows upward through R7 throughout the entire cycle. Likewise, the voltage it develops across R7 is constant rather than fluctuating. The capacitor is able to smooth out the current by delivering extra electrons to the emitter during the negative half-cycles and receiving extra electrons (from the current flowing up through R7) during the positive half-cycles. The filter current which flows between the lower plate of C5 and ground follows the movements of electron flow on and off the upper plate. For instance, extra electrons are drawn into the transistor from the upper plate of C5 during the negative half-cycles, so the filter current flows up from ground into the lower plate. During the positive half-cycles, the upper plate of C5 "recharges" by receiving additional electrons from R7, so the filter current flows down from the lower plate of CS to ground.

Resistor R5 in the voltage-divider circuit is also bypassed with a filter capacitor, C6. This is necessary because the driving currents for the two transistors (both in solid blue) flow through RS in opposite directions. Without this capacitor, the voltage each current would develop across RS would oppose the other voltage. Two separate filter currents flow side by side between the lower plate of capacitor C6 and ground. While the driving current through resistor R1 draws a filter current downward from C6 during negative half-cycles, the driving current through R6 is drawing its own filter current upward from ground into C6.

Summary

There are two electron currents which flow through the aver age transistor and the amount of each is closely regulated by the voltage at the base and emitter terminals. These are the biasing or controlling current, which flows from base to emitter ( or emitter to base in NPN transistors); and the main electron stream (called collector-emitter current, or more commonly, the col lector current), which flows from emitter to collector in NPN transistors.

The amount of biasing current which flows from emitter to base is regulated by the instantaneous voltages at the emitter and base. The voltage at the collector, like the voltage at the plate of a pentode amplifier, has an almost insignificant effect on the amount of collector current which flows.

The voltage at the base of transistor X1 is the algebraic sum of three voltages ( one fixed and two variable) . The fixed voltage is that developed across RS by the voltage-divider current.

The variable voltages are the ones developed across R1-one by the downward flow of base-emitter current, and the other by the signal current.

The instantaneous voltage at the emitter of transistor X1 is the algebraic sum of three voltages, all variable. Two of them are pulsating direct currents and are the two transistor currents--namely, the emitter-base and emitter-collector. The one which is alternating is the feedback current (shown in dotted blue) that flows up and down through resistor R2.

The instantaneous voltage at the base of transistor X2 is the algebraic sum of three independent voltages. The one fixed voltage is developed across resistor RS by the voltage-divider current. One of the two variable voltages is developed across driving resistor R6 by the amplified signal current, which actually flows in two directions through the resistor. The second variable voltage is developed across this same resistor by the emitter-base biasing current (in solid green)--a pulsating direct current which flows downward through R6.

The voltage at the emitter of X2 is the sum of two fixed voltages; therefore, it can probably be described as a "pure" direct voltage. The two fixed voltages which contribute to it are those developed across emitter resistor R7 by the base-emitter and collector-emitter current through the transistor. Both cur rents pulsate through the transistor; but thanks to the filtering action of capacitor C5, they are "pure" DC when flowing through R7.

COMPLEMENTARY SYMMETRY PUSH-PULL AMPLIFIER

Figs. 7, 8, and 9 show three different conditions in the operation of a push-pull amplifier using the complementary symmetry of an NPN and a PNP transistor. This feature permits using the push-pull connection to drive a speaker without the necessity of an output transformer. Both are classed as power transistors, because of the relatively large collector current each must deliver for driving the speaker.

Fig. 7 shows the currents which flow in this circuit under static conditions only (while no signal voltage is being amplified). Fig. 8 has been labeled a negative half-cycle, because the signal current (in solid blue) flows in such directions through driving resistors R1 and R4 that it makes the bases of the two transistors negative. (The bases are their input points.) Fig. 9 has been labeled the positive half-cycle, because the signal current causes positive voltages to exist at the two bases.

Identification of Components

The following components perform the indicated functions in this circuit:

R1-Voltage-divider and biasing resistor.

R2-Voltage-divider and biasing resistor.

R3-Emitter stabilizing resistor.

R4-Voltage-divider and biasing resistor.

R5--Voltage-divider and biasing resistor.

R6-Emitter stabilizing resistor.

C1-Input capacitor.

C2-lnput capacitor.

C3-Emitter bypass capacitor.

C4-Emitter bypass capacitor.

L1-Voice coil of the speaker.

X1-NPN power transistor.

X2-PNP power transistor.

M1-Battery or power supply.

M2-Battery or power supply.

Identification of Currents

During the static period depicted by Fig. 7, no signal currents are flowing in this circuit; but the following electron currents, all pure DC, will flow:

1. Voltage-divider and biasing currents for both transistors (dotted green).

2. Base-emitter current through both transistors (solid green).

3. Collector-emitter current through both transistors, (solid red). During the period of dynamic operation depicted by Figs. 8 and 9, the following additional electron currents will come into existence:

4. Input signal current (solid blue).

5. Fluctuations in the two base-emitter currents ( also in solid green).

6. Fluctuations in the currents through the two voltage-divider circuits (also in dotted green).

7. Fluctuations in the two collector-emitter currents (also in solid red).

8. Current through the speaker voice coil (dotted red)

9. Two emitter filter currents ( dotted red).

Details of Operation

Being fairly straightforward the three electron currents shown in Fig. 7 for the static period of operation can be easily explained. The voltage-divider current (dotted green) flows continuously counterclockwise, upward through the two battery power supplies and to the left through resistor R2. Next it heads downward through resistors R1 and R4, then to the right through resistor R5, where it re-enters the positive terminal of the battery M2.

Electron current will flow around the base-emitter closed loop circuitry of any transistor, in accordance with the voltage biasing conditions at the base and emitter. Transistor X1, an NPN, has an initial negative voltage applied to its emitter from the negative terminal of battery M1. Because of the voltage divider current action at the junction of resistors R1 and R2, however, a less negative voltage is applied to its base.

The voltage polarities around the circuit loop consisting of M1, R2, and R1 may be estimated qualitatively by recognizing that the voltage at the right terminal of resistor R2 will have the same voltage as the battery M1 (-22.5 volts). The voltage at the left terminal of R2 ( this is the voltage applied directly to the base of X1) must be somewhat lower than this value (meaning somewhat less negative) because the voltage-divider current flows to the left through this resistor. Recall that the terminal from which electron current leaves a resistor is always more positive (or less negative) than the terminal at which it enters the resistor.


Fig. 7. Current conditions in the complementary symmetry push-pull amplifier-static operation with no signal currents.

Since the junction of resistors R1 and R4 is connected directly to ground, the voltage at this point must always be zero. The junction of the two batteries is also grounded. As the voltage divider current continues around this closed loop, it moves into a region of progressively higher positive voltages. The junction of R4 and R5 is positive with respect to ground, and the right terminal of R5 will always be at the full +22.5 volts of the battery.

Under no-signal conditions, the base-emitter current for both transistors (solid green) also flows continuously in a counter clockwise loop. Its path might be considered to begin at the ground point between the two batteries. From here it flows upward through M1 and resistor R3, into the emitter and out the base of XL Heading downward through R1 and R4, it goes into the base and out the emitter of transistor X2, then through R6 and into the positive terminal of M2. Here, it flows through M2 back to the reference point or reference voltage which we call ground.

This base-emitter current of a transistor is usually called the "biasing" current, and it controls closely the amount of collector current (meaning collector-emitter current) which flows through the transistor. This phenomenon is the basis for the descriptive statement that transistors are "current-controlled" devices-in contrast to vacuum tubes, which are considered to be "voltage-controlled" devices.

The amount of this base-emitter biasing current is itself closely controlled and regulated by the voltage values at the base and emitter. Under no-signal conditions such as are depicted in Fig. 7, the voltage at the base of X1 is negative. In fact, it is the sum of (1) the negative voltage produced at the junction of R1 and R2 by the voltage divider current, and (2) the negative voltage caused at the upper end of resistor R1, by the base emitter current flowing downward through R1.

The latter voltage may be more easily visualized by recognizing the current path which includes battery M1, resistor R3, the resistance of the junction between emitter and base within the transistor, and resistor R1 as another voltage divider. Starting at the negative terminal of battery M1, the voltage has its maximum negative value. Proceeding counterclockwise around this loop, the voltage at each point becomes progressively less negative until the ground point at the lower end of resistor R1 is reached, where the voltage is zero.

Under the assumption that the two transistors will conduct identical amounts of biasing current during this no-signal condition, the same amount of base-emitter current (in solid green) continues to flow through resistor R4, the resistance of the junction between base and emitter within transistor X1, resistor R6, and battery M2. This loop constitutes another voltage divider, the voltage again becoming progressively more positive (i.e., less negative) as this electron current proceeds around the loop. Electrons move in the direction indicated because they flow inescapably away from more negative and toward more positive voltages.

Thus the "biasing" voltage at the base of transistor X2 is positive and is the sum of two separate components of positive voltage, both produced across resistor R4 by the two currents which flow downward through it.

The voltage at the emitter of transistor X1 is negative and is the sum of two separate components of negative voltage. These are produced at the upper terminal of resistor R3 by the two electron currents (the base-emitter current, in solid green; and the collector-emitter current, in solid red) flowing upward through R3.

The voltage at the emitter of transistor X2 is positive, and is the sum of two separate positive voltages. Both exist at the lower terminal of resistor R6 as a result of the same two electron currents-the base-emitter and collector-emitter currents which flow upward through it.

The path for the combined collector currents (under these no-signal conditions) might be considered to start at ground (the reference point) between the two batteries, and to flow up ward through M1 and R3 into the emitter and out the collector of X1. Next it heads downward to the collector of X1, through X1 and out the emitter, then upward through R6 to the positive terminal of M2. This electron current is finally returned to ground by flowing through M2 to its negative terminal.

Operation Under Dynamic Conditions

Once these three static currents have been visualized, you will be in a much better position to understand how this circuit operates when a signal is being amplified. Three additional currents come into existence. Also, the three currents previously discussed will now be changed from pure to pulsating DC. These pulsations represent the variations in the audio-frequency signal.

During the negative half-cycle depicted by Fig. 8, a signal voltage of negative polarity is applied to the left plates of capacitors C1 and C2. This negative voltage drives electron current (in solid blue) onto the left plates of these capacitors.

In turn equal amounts of electron currents are driven away from the right plates. The current driven away from C1 flows down ward through resistor R1 to ground and, in so doing, develops an instantaneous component of negative voltage at the upper terminal of R1. This component of negative voltage adds to the negative voltage already existing at that point as a result of the two currents which flow through R1. The increase in negative voltage at the base of X1 ( or any other NPN transistor) will decrease the base-emitter biasing current flowing through the transistor. This, in turn, decreases the collector-emitter current through X1.

Unlike the collector current through transistor X1 the collector current through transistor X2 increases during negative half-cycles. The reasons why it does may be summarized briefly, as follows:

1. The input signal current (in solid blue) flows upward through resistor R4 to ground. The component of negative voltage it creates at the lower terminal of R4 must be subtracted from the positive voltage already existing at that point as a result of the two voltage-divider actions previously described. The net result is a reduction in the positive voltage at the base of transistor X1.

2. A reduction in positive voltage at the base of any PNP transistor will always increase the flow of biasing current.

Therefore, more base-emitter current flows through transistor X1.

3. The increase in biasing current causes a companion in crease in collector-emitter current, which is also the load current.


Fig. 8. Operation of the complementary symmetry push-pull amplifier negative half-cycle.

It should be obvious that this additional collector current for X2 cannot all be drawn through X1, since the collector current through X1 is decreasing during this negative half-cycle, as previously explained. Consequently, the extra collector current for X2 is drawn upward from ground and through the voice coil of the speaker. This current (in dotted red) will set up magnetic lines of force which will draw the speaker diaphragm in one direction only, such as to the right in Fig. 8. The complete path of this current is through the voice coil, then into the collector and out the emitter of X2 upward through resistor R6 to the positive terminal of battery M2. Flowing through the battery to its negative terminal, the current re-enters the common ground, which provides a ready return access to the voice coil.

During the positive half-cycles depicted by Fig. 9, a similarly long series of events (which will be discussed later) increases the collector current through X1 and decreases the collector current through X2. This increase in collector cur rent through X1 cannot be accepted by X2 during this same time period. Therefore, the current is driven through the speaker voice coil and into the ground connection. Thus, current flows through the voice coil in one direction during negative half cycles, and in the opposite direction during positive half-cycles.

This causes the resultant magnetic lines of force in the voice coil to change direction every half-cycle. As a result, the permanent magnet (and speaker diaphragm connected to it) will be alternately attracted to and repelled by the voice coil during each cycle. The back-and-forth movements of this diaphragm set up the air vibrations we know as sound waves.

To consider all the current and voltage changes which occur during a positive half-cycle of operation, we should start with the input signal. As long as its polarity is positive, electrons will be drawn onto the right plates of C1 and C2. As it flows upward through resistor R1 and downward through resistor R4, this electron current places small components of positive voltage at the upper terminal of R1 and the lower terminal of R4. These voltage components are added to the voltage already existing there as a result of the two voltage-divider actions previously discussed (the ones associated with the currents in solid and dotted green). This signal current flow through the two resistors makes the normally negative voltage at the top of R1 less negative, and the normally positive voltage at the lower terminal of R4 more positive. As a result, the two currents flowing through transistor X1 ( the base-emitter biasing current and the collector-emitter load current) increase, and the two flowing through transistor X2 decrease.

Thus, by application of a small audio-frequency signal at the input point of this circuit, it is possible to cause a fairly heavy flow of alternating current at the same frequency through the speaker voice coil. This circuit is classed as a power amplifier because much more audio-frequency power is delivered to the output (speaker) than is consumed in the input circuit (resistors R1 and R4). As an example, the speaker might be delivering one or more watts of audio power, whereas only a few milliwatts are being consumed in input resistors R1 and R4.

Several important facts about transistors make this phenomenon possible:

1. In a transistor, a change of only a fraction of a volt in the voltage difference between base and emitter will cause a small change in the amount of biasing current and a much larger change in the amount of collector load current.

2. This fraction of a volt required can be developed at the transistor base by causing a current as small as a fraction of a milliampere to flow through an input resistor such as R1 or R4.

Moreover, the collector-emitter load current flowing through these power transistors may be hundreds of times greater than the signal current flowing up and down through input resistors R1 and R4.

Because of the voltage changes across these two resistors while the signal current is flowing, the amount of voltage-divider current (in dotted green) will be disturbed slightly. During the negative half-cycles (Fig. 8), the voltage at the junction of R1 and R2 is made more negative; consequently, the smaller voltage difference between the two terminals of resistor R2 will draw less current through R2 from the negative terminal of the battery.


Fig. 9. Operation of the complementary symmetry push-pull amplifier positive half-cycle.

During the positive half-cycles, this junction is made less negative and now the opposite is true-a larger voltage-divider current will be driven through R2 by the greater voltage difference across its terminals.

By similar reasoning, it can be shown that the voltage-divider current flowing through the lower network ( consisting of R4, R5, and M2) also fluctuates with the applied signal voltage. Here, however, it increases during the negative and decreases during the positive half-cycles.

For this reason, the voltage-divider current (in dotted green) has been shown as two separate currents-flowing in two separate networks-while a signal is applied. Any difference in amount between these two currents will automatically be compensated for by a flow of electron current through the common ground connection between the junction of power sources M1 and M2, and the junction of resistors R1 and R4.

Likewise, any difference in amounts between the base-emitter currents through the two transistors will be compensated for by an appropriate flow of electron current (in solid green) through this same common ground connection. During the positive half cycles (when more base-emitter current flows through X1 than through X2), this current will flow through ground from left to right-that is, from the junction of resistors R1 and R4, toward the junction between the two batteries. Conversely, it will flow through ground from right to left during the negative half-cycles, when the base-emitter current through X2 exceeds that through X1.

Filter Currents

Emitter resistors R3 and R6 are bypassed by filter capacitors C3 and C4, respectively, to prevent loss of signal strength from that type of negative feedback known as degeneration. In order to forestall degeneration, the filter currents shown in dotted red must be able to flow freely between C3 and C4 to the nearest ground point. The voltage on the upper plate of C3 must always be identical to the one at the emitter of X1. Since this is a negative voltage, the accumulated charge on the upper plate of capacitor C3 can be conveniently represented as a "pool" of electrons. There will be a continual flow of electron current driven upward through resistor R3 by the negative terminal of battery M1. (This is the point of most negative voltage in the entire circuit.) If no collector current could escape into the transistor, the upper plate of capacitor C3 would soon acquire enough electrons that its voltage would equal the full negative battery voltage. At that time, the upward flow of electron current through R3 would cease. However, since transistor X1 is being operated under what are called Class-A conditions, some collector current flows throughout the entire cycle. Therefore the negative voltage on the upper plate of C3 will never attain the full negative voltage power source M1.

During the positive half-cycle such as in Fig. 9, the biasing conditions of transistor X1 cause an increase in the collector current. In turn, the electrons which make up this current are drawn quite easily from the electron pool on the upper plate of capacitor C3, and an equal number flows upward from ground to the lower plate. This is the essence of capacitor action . . . it is how capacitors will appear to "pass" an alternating current.

When the capacitor has sufficient size, or capacity, the quantity of extra electrons demanded by the transistor collector current during the positive half-cycles is such an infinitesimal percentage of the total number of electrons stored in the capacitor that the voltage does not change appreciably from half-cycle to half-cycle.

This is how a filter capacitor avoids the loss in signal strength known as degeneration. If emitter resistor R3 were not bypassed with a filter capacitor, then the increased collector current during the positive half-cycles would have to be drawn directly upward through R3. This would make the voltage difference, or "drop," across R3 larger, and would lower the negative voltage at the upper terminal of R3. Since the emitter of X1 is connected directly to this point, this lower negative voltage would constitute a change in the biasing conditions of the transistor, and would ultimately reduce the collector current flowing through the transistor. This would nullify at least part of the original increase in collector current, and would constitute a loss in the amount of amplification the circuit delivers. This is degeneration.

The voltage on the lower plate of capacitor C4 is always identical to the positive voltage at the emitter of transistor X1, since they are connected together. It is convenient to visualize a positive capacitor voltage as a pool of positive ions, out of which a continual flow of electrons will move upward and through resistor R6, toward the higher positive voltage of power source M2. If no collector-emitter current flowed through X2, the positive voltage on the lower plate of C4 would eventually equal the +22.5 volts of the battery at the lower terminal of M2. Instead, the continual inflow of collector-emitter current from X2 keeps it at a lower positive value.

Assuming capacitor C4 has sufficient size, or capacity, the quantity of excess electrons which flow onto its lower plate from the collector-emitter current during the negative half-cycles (Fig. 8) is such an infinitesimal percentage of the total number of positive ions already stored there that the voltage across capacitor C4 does not change appreciably from half-cycle to half cycle.

If C4 were not in the circuit, or if for any reason the filter current between it and ground were prevented from flowing, then degeneration (loss of signal strength) would occur across R6.

The increased collector-emitter current through X2 during negative half-cycles would flow directly through R6 and thereby increase the voltage drop across this resistor. The lower positive voltage now produced at its lower terminal and at the emitter would be a fundamental change in the biasing conditions of the transistor. The resultant decrease in collector-emitter current would nullify part of the original increase in this current and thereby lead to some loss in amplification.


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